alexepascual said:
I agree. Now the question is: When I write the expression for the phase as function of time in the lower arm, I need to know what energy to use. On one hand I have read that once the particles are entangled you can't consider the separate Hilbert spaces anymore. In that case I would like to use the sum of the energies of both particles, even if I am looking at only one particle there. On the other hand I couls say that because I am considering only one particle in the lower arm, I should consider only the energy for that particle. In one case the frequency is twice what it would be in the other case. So, even if I believe that the wave funtion only exists in my brain, I need numbers to plug in for p and E.
I'm still not quite clear on what you're asking. The E and |p| are as you would expect and the same as would be defined classically. You just won't be able to write down a wave-function where E reflects the frequency.
(Below I am using A and B to indicate the A particle and B particle of an entangled pair)
The proper way to talk about one half of an entangled pair is with the reduced density operator. In the continuous representation (i.e. with wave-functions) this will take the form of a function of two positions:
\rho_A(\vec{x},\vec{y})
(also a function of time) (Note this is another doubling of the variables)
Think of the vectors x and y as row and column indices for a matrix.
The bifunction forms act as operators on functions via:
T[f]=g \quad \equiv \quad g(x) = \int T(x,y)f(y)dy
The full density operator for the entangled pair will then have four vector variables:
\rho_{AB}(\vec{x}_A,\vec{y}_A,\vec{x}_B,\vec{y}_B)
and the reduced density operator is obtained by "tracing over" the variables corresponding to the other half of the pair:
\rho_A(\vec{x},\vec{y}) = \iint \delta^3(\vec{x}_B-\vec{y}_B)\rho_{AB}(\vec{x}_A,\vec{y}_A,\vec{x}_B,\vec{y}_B)d^3x_B d^3 y_B
In the case of a sharp mode (for the whole entangled pair) this density operator will take the form:
\rho_{AB}( = \psi(x_A,x_B)\psi^\dag(y_A,y_B)
(henceforth assume x's and y's are vectors.)
But once you take the partial trace to get the reduced density operator you won't have such a factorable form. (It is not in a sharp mode)
Now I get to the point... At best you can write the reduced density operator as a
(classical) probability weighted average of such sharp density operators.
\rho_A(x,y) = \sum_{k}p_k\rho_k(x,y) =\sum_k p_k \psi_k(x)\psi^\dag_k(y)
Each such will be a product of wave-functions as above where each of those wave-functions has the same energy and thus frequency.
\psi_k(x) = \phi_k(x)e^{i\omega t}
You can't speak meaningfully about the "phase of the lower (A) particle" at a given point or even at a given time. When you consider only one half of an entangled pair you get something similar to a particle from a noisy source and you must work within the density operator format.
And this is my point about being "a good Copenhagenist". Get out of the habit of thinking in terms of "the phase of a particle". Its the wave-function which has a phase and without a wave-function (since the mode is not a sharp one) we can't define phase.
As the ole codger on the porch says... "Ya caint get there from here!"
Oh! by the way.. Do you use some editor for LateX?
I use both Led and WinShell (both free). Sometimes I prefer one sometimes I prefer the other. Right now Led is the main one I use.
